If \(\sqrt{2}\) were rational, then \(\sqrt{2}=\frac{p}{q}\text{,}\) with \(p\) and \(q\) coprime. But then \(2q^2=p^2\text{.}\) By the Fundamental Theorem of Arithmetic, the power of \(2\) dividing the left side is odd, while the power of \(2\) dividing the right side is even. This is a contradiction, so \(\sqrt{2}\) is not rational.